Answer to Question #107180 in Analytic Geometry for Nikesh gautam pandit ji

Question

Show that the closed sphere with centre (2,3,7)and radius 10 in R^3 is contained in the open cube P = {(x, y,z): |x − 2 |<11, |y − 3| <11, |z − 7| <11}.

Answer 1

Equation of the sphere with centre (2,3,7) and radius 10 in $R^3 \implies (x-2)^2+(y-3)^2+(z-7)^2=10^2$

Thus, extremities of the sphere in all 3 directions can be found as :

$|x-2| <10, |y − 3| <10, |z − 7| <10$ .

But the given cube is $P = \{(x, y,z): |x − 2 |<11, |y − 3| <11, |z − 7| <11\}$

Thus, clearly the sphere lies inside the cube as its extremities are smaller than extremities of the cube!